mathop {lim }limits_{x to 0} frac{{1 - cos mx | vedclass

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(C) Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \matho

Vedclass · Accepted Answer

$\frac{m^2}{n^2}$

Vedclass · Answer

(C) Using the trigonometric identity $1 - \cos 	heta = 2 \sin^2(	heta/2)$:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x 	o 0} \frac{2 \sin^2(mx/2)}{2 \sin^2(nx/2)}$$= \mathop {\lim }\limits_{x 	o 0} \left[ \left( \frac{\sin(mx/2)}{mx/2} ight)^2 \cdot \frac{m^2 x^2}{4} \cdot \left( \frac{nx/2}{\sin(nx/2)} ight)^2 \cdot \frac{4}{n^2 x^2} ight]$Since $\mathop {\lim }\limits_{	heta 	o 0} \frac{\sin 	heta}{	heta} = 1$,we get:$= \frac{m^2}{n^2} \cdot 1 = \frac{m^2}{n^2}$.Alternative method: Apply $L$-Hospital's rule:$\mathop {\lim }\limits_{x 	o 0} \frac{1 - \cos mx}{1 - \cos nx} = \mathop {\lim }\limits_{x 	o 0} \frac{m \sin mx}{n \sin nx}$$= \mathop {\lim }\limits_{x 	o 0} \frac{m^2 \cos mx}{n^2 \cos nx} = \frac{m^2(1)}{n^2(1)} = \frac{m^2}{n^2}$.

$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \cos mx}}{{1 - \cos nx}} = $

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